How would you solve i^27? thanks so much!!

Im really confused on this! i also have to solve -3i^422 and 3i+5i^41
Thanks SO much if anyone can explain this!! It would be a tremendous help!!

The number i follows a cycle.

i = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
remember that i = sqrt(-1). All you need to do is perform a mod 4 function on the exponent, and use that to determine your position in the cycle. 27 mod 4 = 3, which means it's really i^3, which is -i.

For -3i^422, we get this: 422 mod 4 = 2, -3i^2, -3*-1, 3
For 3i + 5i^41, we get: 41 mod 4 = 1, 3i + 5i, 8i

i = √(-1), so i² = -1
If we have i^4 = i² i² = -1 * -1 = 1

Therefore, we can just ignore every term of i^4.

i^27 = i^(27 mod 4) = i^3 = i² i = -1 * i = -i.

i^2= -1 right ? ( or i^4= 1 )

i ^27 = i^26 *i
= (i^2)^13 * i
= (-1) ^13 *i
= -1*i
= -i

-3i^422 = -3(i^2)^( 211) = -3 (-1)^211 = -3 (-1) = +3

i^41 = i^40 * i= (i^2)^20 *i = (-1)^20 *i = (1) *i = i
3i+5i^41 = 3i+i =4i

Recall that i^2 = -1, i^3 = -i, i^4 = 1

i^27 = i^24*i^3 = (i^4)^6 * i^3 = 1^6*(-i) = -i

-3 i^422 = -3* i^420* i^2 = -3* (i^4)^105 * i^2 = -3* 1^5*(-1) = 3

3i + 5i^41 = 3i + 5 (i^40)i = 3i + 5i = 8i